What Must Be Checked
A collection of subsets can cover a space and have well-behaved intersections without being a basis for a particular topology on that space. The key is to keep two requirements separate: basis elements must be open in the topology being considered, and they must provide sufficiently small building blocks inside every open set. The examples below test these requirements on familiar intervals and small finite sets.
Let \((X,\mathcal{T})\) be a topological space. A collection \(\mathcal{B}\) is a basis for \(\mathcal{T}\) if every member of \(\mathcal{B}\) is open and every open set is a union of members of \(\mathcal{B}\). The union of an empty collection is understood to be \(\varnothing\). In particular, the definition requires \(\mathcal{B}\subseteq\mathcal{T}\); representing open sets as unions is not enough if some proposed building blocks are not open.
Proof. Suppose first that \(\mathcal{B}\) is a basis. Its members are open by definition. If \(U\) is open, write \(U\) as a union of members of \(\mathcal{B}\). For any \(x\in U\), at least one set in that union contains \(x\); call it \(B\). Since the union equals \(U\), this gives \(x\in B\subseteq U\).
Conversely, suppose both conditions hold. Let \(U\in\mathcal{T}\). For every \(x\in U\), condition (2) supplies a member \(B_x\in\mathcal{B}\) with \(x\in B_x\subseteq U\). Thus \(U\) is contained in the union of all members of \(\mathcal{B}\) that are subsets of \(U\). That union is also contained in \(U\), so the two sets are equal. If \(U=\varnothing\), it is the empty union. Every open set is therefore a union of members of \(\mathcal{B}\), and condition (1) says those members are open. Hence \(\mathcal{B}\) is a basis for \(\mathcal{T}\). \(\square\)
This test gives a direct way to check a candidate collection: first check that its members are open; then test whether every point in every open set has a candidate basis element around it that stays inside that open set. A related overlap condition is a useful necessary check.
Proof. Since \(B_1\) and \(B_2\) are open, their intersection \(B_1\cap B_2\) is open. It contains \(x\). Apply the local-refinement test to this open set and point: there is \(B_3\in\mathcal{B}\) with \(x\in B_3\subseteq B_1\cap B_2\), as required. \(\square\)
The overlap condition detects a common failure: two proposed basis elements overlap, but there is no candidate element small enough to fit around a point of that overlap. Passing this condition alone does not prove that a collection is a basis for a specified topology. The elements must still be open, and local refinement must hold for every open set in that topology.
Open Intervals in the Standard Real Line
Worked Example: Bounded Open Intervals
Let \(\mathbb{R}\) have its standard topology, and let \(\mathcal{B}\) be the collection of all bounded open intervals \((a,b)\) with \(a,b\in\mathbb{R}\) and \(a<b\). We check that \(\mathcal{B}\) is a basis.
Openness. Each member \((a,b)\) is open in the standard topology. Indeed, if \(x\in(a,b)\), set \(r=\min(x-a,b-x)/2\). Both \(x-a\) and \(b-x\) are positive, so \(r>0\); also \(x-r>a\) and \(x+r<b\). Therefore the open interval \((x-r,x+r)\) contains \(x\) and is contained in \((a,b)\).
Local refinement. Let \(U\) be open in the standard topology and let \(x\in U\). By the definition of this topology, there is an \(r>0\) such that \((x-r,x+r)\subseteq U\). The bounded open interval \(B=(x-r/2,x+r/2)\) belongs to \(\mathcal{B}\), contains \(x\), and is contained in \(U\). Thus the local-refinement test applies.
Overlap check. Suppose \((a,b)\) and \((c,d)\) both contain \(x\). Then \(a<x<b\) and \(c<x<d\). Set \(L=\max(a,c)\) and \(R=\min(b,d)\). These inequalities give \(L<x<R\), so \(L<R\), and \((L,R)\) is a bounded open interval containing \(x\) and contained in \((a,b)\cap(c,d)\).
Check: For \(U=(-2,5)\) and \(x=1\), take \(r=1\), since \((0,2)\subseteq(-2,5)\). The chosen basis element is \((1-1/2,1+1/2)=(1/2,3/2)\); it contains \(1\) and satisfies \((1/2,3/2)\subseteq(-2,5)\).
Overlapping Subsets on a Finite Space
Worked Example: An Overlap That Has a Basis Element
Let \(X=\{a,b,c\}\) and give it the topology \(\mathcal{T}=\{\varnothing,\{a\},\{a,b\},\{a,c\},X\}\). Take \(\mathcal{B}=\{\{a\},\{a,b\},\{a,c\},X\}\). We verify that this is a basis for this specified topology.
Every member of \(\mathcal{B}\) appears in \(\mathcal{T}\), so each is open. To check local refinement, let \(U\in\mathcal{T}\) and \(x\in U\). Since every nonempty member of \(\mathcal{T}\) is itself in \(\mathcal{B}\), choose \(B=U\). Then \(x\in B\subseteq U\). The condition is vacuous for \(U=\varnothing\). The local-refinement test proves that \(\mathcal{B}\) is a basis.
The overlap of \(\{a,b\}\) and \(\{a,c\}\) is \(\{a\}\). At the overlap point \(a\), the basis element \(\{a\}\) satisfies \(a\in\{a\}\subseteq\{a,b\}\cap\{a,c\}\). Other nonempty intersections are also covered: \(\{a\}\) intersected with any member containing \(a\) gives \(\{a\}\), and the intersection of \(X\) with any member is that member.
Check: For the open set \(U=\{a,c\}\) and the point \(x=c\), choose \(B=\{a,c\}\). Then \(c\in\{a,c\}\subseteq\{a,c\}\). For \(x=a\) in the same open set, choose \(B=\{a\}\); this also contains \(a\) and is contained in \(U\).
Worked Example: An Overlap With No Small Enough Element
Let \(X=\{p,q,r\}\) have the discrete topology \(\mathcal{T}=\mathcal{P}(X)\), so every subset is open. Consider \(\mathcal{B}=\{X,\{p,q\},\{q,r\}\}\). These sets cover \(X\), and every candidate is open, but the collection is not a basis.
The two members \(\{p,q\}\) and \(\{q,r\}\) overlap at \(q\), and their intersection is \(\{q\}\). The only member of \(\mathcal{B}\) containing \(q\) is either \(X\), \(\{p,q\}\), or \(\{q,r\}\). None is contained in \(\{q\}\). Thus the overlap condition fails. Equivalently, the open set \(U=\{q\}\) and point \(q\in U\) have no candidate \(B\) with \(q\in B\subseteq U\). The local-refinement test confirms that \(\mathcal{B}\) is not a basis.
Check: The failure is not a lack of coverage: \(p\in\{p,q\}\), \(q\in\{p,q\}\cap\{q,r\}\), and \(r\in\{q,r\}\). Nor is it a lack of openness, because the topology is discrete. The exact failure is refinement at the open singleton \(\{q\}\).
Singletons: Refinement Is Not Openness
Worked Example: Singletons in the Standard Topology
Let \(X=\mathbb{R}\) with its standard topology and let \(\mathcal{S}=\{\{x\}:x\in\mathbb{R}\}\). This collection covers \(\mathbb{R}\), and for every open \(U\) and every \(x\in U\), the member \(\{x\}\) satisfies \(x\in\{x\}\subseteq U\). Nevertheless, \(\mathcal{S}\) is not a basis for the standard topology, because its members are not open.
To verify that \(\{x\}\) is not open, take any \(r>0\). The interval \((x-r,x+r)\) contains \(x+r/2\), and \(x+r/2\ne x\). Consequently no open interval around \(x\) is contained in \(\{x\}\). This proves that \(\{x\}\) is not open in the standard topology. Since every member of a basis for a specified topology must be open, \(\mathcal{S}\) fails that requirement.
Check: For \(U=(0,1)\) and \(x=1/2\), the local-refinement condition does hold: \(1/2\in\{1/2\}\subseteq(0,1)\). The collection still fails to be a basis because \(\{1/2\}\notin\mathcal{T}\). In the discrete topology on any set \(X\), by contrast, every singleton is open, and the singleton collection is a basis: it covers \(X\), and for every \(U\subseteq X\) and \(x\in U\), \(\{x\}\) is a member contained in \(U\).
A Reliable Checking Order
A frequent mistake is to confuse the local-refinement test with the requirement that the proposed sets themselves be open. For instance, one might incorrectly reject the singleton \(\{1/2\}\) as a local refinement of \((0,1)\), on the grounds that it contains no interval around \(1/2\). That calculation tests the wrong condition: local refinement asks whether \(1/2\in\{1/2\}\subseteq(0,1)\), which is true. The actual failure is that \(\{1/2\}\) is not open in the standard topology.
Another tempting shortcut is to check only that candidate sets cover the space and that intersections can be refined. The discrete-space example shows why this does not establish a basis for a specified topology: even though the candidates are open and cover \(X\), their overlap at \(q\) cannot be refined inside the open set \(\{q\}\). Conversely, passing an overlap check does not excuse a failure of openness, as the singleton collection on \(\mathbb{R}\) illustrates.
For a candidate collection on a specified space, check the conditions in this order:
- Verify that every proposed basis element is open in the stated topology.
- For an arbitrary open \(U\) and point \(x\in U\), find a candidate \(B\) with \(x\in B\subseteq U\).
- Use overlaps of candidate sets as a focused test: at every point of an overlap, look for a candidate contained in that intersection.
The third check is especially effective for finding failures, while the first two together give the full test. If \(X=\varnothing\), the local condition has no points to test: the empty collection is a basis for the unique topology \(\{\varnothing\}\), since its only open set is the empty union.
Check Your Understanding
For each question, specify the topology being used and distinguish openness from local refinement.
- In the standard topology on \(\mathbb{R}\), test the collection of all bounded open intervals with rational endpoints using the local-refinement test.
- On \(X=\{a,b,c\}\) with the discrete topology, decide whether \(\{\{a,b\},\{b,c\},\{a\},\{c\}\}\) is a basis. Check the open set \(\{b\}\) carefully.
- In the standard topology on \(\mathbb{R}\), do all closed bounded intervals form a basis? Identify which condition succeeds or fails.
- Explain why the overlap condition is necessary for a basis, and why satisfying it alone does not prove that a collection is a basis for a specified topology.
- For the discrete topology on a set \(X\), verify directly that the collection of all singletons is a basis, including when \(X=\varnothing\).